If the image of the point P(1,2,a) in the line (x-6)/(3)=(y-7)/(2)=(7-z)/(2) is Q(5,b,c), …

Question

If the image of the point $P(1,2,a)$ in the line $\frac{x-6}{3}=\frac{y-7}{2}=\frac{7-z}{2}$ is $Q(5,b,c)$, then ${a}^{2}+{b}^{2}+{c}^{2}$ is equal to

Accepted Answer

Direction of the line is $(3,2,-2)$ (writing $\dfrac{7-z}{2}=\dfrac{z-7}{-2}$). The midpoint of $PQ$, $M=\left(3,\ \dfrac{2+b}{2},\ \dfrac{a+c}{2}\right)$, lies on the line, and $\vec{PQ}=(4,\ b-2,\ c-a)$ is perpendicular to $(3,2,-2)$.

From $M$ on the line: $\dfrac{3-6}{3}=-1 \Rightarrow \dfrac{y-7}{2}=-1\Rightarrow y=5\Rightarrow \dfrac{2+b}{2}=5\Rightarrow b=8$; and $\dfrac{7-z}{2}=-1\Rightarrow z=9 \Rightarrow \dfrac{a+c}{2}=9 \Rightarrow a+c=18.$

From perpendicularity: $3(4)+2(b-2)-2(c-a)=0 \Rightarrow 12+2b-4-2c+2a=0$. With $b=8$: $2a-2c+24=0 \Rightarrow a-c=-12.$

Solving $a+c=18,\ a-c=-12$: $a=3,\ c=15$.
$a^2+b^2+c^2=9+64+225=298.$

Solutions · 1